[^(-1)\(^(-1)x)\] is equal to Option 1: √((x^2+1)/(x^2+2)) Option 2: √((x^2+2)/(x^2+1)) Option 3: √(x/(x^2+2)) Option 4: √(x/(x^2+1))

Answer: √((x^2+1)/(x^2+2)). Solution: Let =^(-1)x, so =x and =1/(√(1+x^2)). Now let =^(-1)1/(√(1+x^2)). Then the hypotenuse is √(1+(1+x^2))=√(x^2+2). Hence =(√(x^2+1))/(√(x^2+2))=√((x^2+1)/(x^2+2)).

Geometry

Trigonometry

Medium

core

$\cos[\tan^{-1}\{\sin(\cot^{-1}x)\}]$ is equal to

$\sqrt{\frac{x^2+1}{x^2+2}}$

$\sqrt{\frac{x^2+2}{x^2+1}}$

$\sqrt{\frac{x}{x^2+2}}$

$\sqrt{\frac{x}{x^2+1}}$

✅ Correct Option: 1

Let $\theta=\cot^{-1}x$ , so $\cot\theta=x$ and $\sin\theta=\frac{1}{\sqrt{1+x^2}}$ .

Now let $\phi=\tan^{-1}\frac{1}{\sqrt{1+x^2}}$ . Then the hypotenuse is $\sqrt{1+(1+x^2)}=\sqrt{x^2+2}$ .

Hence $\cos\phi=\frac{\sqrt{x^2+1}}{\sqrt{x^2+2}}=\sqrt{\frac{x^2+1}{x^2+2}}$ .

2025: 15 May Shift 1

For the principal value branch, the value of $\sin\left(\frac{\pi}{2} - \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)$ is

2025: 29 May Shift 2

The value of $\cos(2\cos^{-1}x + \sin^{-1}x)$ at $x = \frac{1}{5}$ is

2025: 19 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\cos^{-1} x + \cos^{-1}(-x)$	(I) $\frac{\pi}{3}$
(B) $\text{cosec}^{-1}(-x) + \sec^{-1}(-x)$	(II) $-\frac{\pi}{3}$
(C) $\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$	(III) $\pi$
(D) $\tan^{-1}\left(\tan\frac{4\pi}{3}\right)$	(IV) $\frac{\pi}{2}$

Choose the correct answer from the options given below:

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The value of $\alpha$ is

2026: 29 May Shift 1

The value of $\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$ is

2025: 13 May Shift 1

The value of $\tan^{-1}(1) + \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right)$ is

cos⁡[tan⁡−1{sin⁡(cot⁡−1x)}]\cos[\tan^{-1}\{\sin(\cot^{-1}x)\}]cos[tan−1{sin(cot−1x)}] is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\cos[\tan^{-1}\{\sin(\cot^{-1}x)\}]$ is equal to